Theorem: C5: uniform-angle coherent amplification

Assume (eq:exactdecouple) and access to the two reflections above. Put

After $r$ iterations of

followed by the final Bell test, the success probability is

The successful recovered map is $s_r{\mathrm{id}}_M$.

If the success flag is forgotten while the failure quantum output is retained, the complete recovered channel has error at most $1-s_r$.

For $k=1$, one iteration gives $s_1=1$, hence deterministic exact recovery. For general $k\ge 1$, choosing a nonnegative integer $r$ nearest to

gives $1-s_r\le m^{-2}$ and $r=O(m)$.

Proof. For a unit vector $v\in \mathrm{supp}S$, let $a_v=Tv$. Equation (eq:Lnorm) gives

Define

These are orthonormal, and

The two reflections preserve their span and their product rotates the angle by $2\theta$. The same angle applies to every $v$ in the encoded support. Therefore the success component after $r$ iterations is

Using Theorem thm:HPpost gives success map $s_r{\mathrm{id}}_M$, including arbitrary references.

The failure map has constant trace $1-s_r$. When this is nonzero, divide it by that scalar to obtain a CPTP channel ${\mathcal{F}}_r$. The complete output channel is

Half-diamond distance between channels is at most one, giving the error bound.

When $m=2$, $\theta =\pi /6$, and $r=1$ gives angle $\pi /2$. For the nearest-integer choice, the final angle differs from $\pi /2$ by at most $\theta$. Hence failure is at most ${\sin }^2\theta =m^{-2}$.